$≥\dfrac{(\sqrt[]x+\sqrt[]y+\sqrt[]z)^2}{\sqrt[]x+\sqrt[]y+\sqrt[]z}=\sqrt[]x+\sqrt[]y+\sqrt[]z$
$(\sqrt[]x+\sqrt[]y+\sqrt[]z)^2≥3(\sqrt[]{xy}+\sqrt[]{yz}+\sqrt[]{zx})=3$
$⇒\sqrt[]x+\sqrt[]y+\sqrt[]z≥\sqrt[]3$
Nên $P≥\sqrt[]3$
Dấu $=$ xảy ra $⇔x=y=z=\dfrac{1}{3}$